# Finding Recoil Velocity

## Homework Statement

Oliver has a mass of 59.1 kg and can throw a 558.0 g rock with a speed of 19.8 m/s. What would Oliver's recoil speed be if he were on an icy surface?

## Homework Equations

Momentum is conserved in an isolated system (when Fnet=0 or no external forces are acting on the system)
p=mv

mass(object 1)*initial velocity (object 1) + mass(object2)*initial velocity (object 2)= Mass (object 1)*final velocity (object 1) + mass (object 2)*final velocity (object 2)

## The Attempt at a Solution

:surprised

m (oliver)= 59.1 kg, m(rock)= 0.588 kg, initial velocity = 19.8 m/s

p(oliver+ rock)= mv
p= (59.1+0.588)(19.8) =1181.8 kg m/s (seems quite too large--> unreasonable?) on ice (frictionless i assume) no net external forces so Fnet=0
i don't understand what recoil velocity means? Is it the rebound velocity? would be equal and opposite?:grumpy:

***Any help would be appreciated **** ## The Attempt at a Solution

Related Introductory Physics Homework Help News on Phys.org
Doc Al
Mentor
Realize that Oliver and his rock are originally at rest on the ice. So what is the original momentum of the sytem?

Also, "recoil velocity" is the speed that Oliver will move back with (recoil) after he throws that rock. (If he throws the rock East, he will recoil to the West.)

p(oliver+ rock)= mv
p= (59.1+0.588)(19.8) =1181.8 kg m/s (seems quite too large--> unreasonable?)
Where is it mentioned that Oliver is travelling at 19.8m/s?

Recoil velocity refers to velocity with which Oliver would be moving AFTER he throws the ball.

Take a look at the equation you posted and just substitute the values.

Would the orginal momentum be zero
so would i use 0 as the momentum and find velocity from there?
i am so confused also it says that if oliver throws the rock he would throw it at 19.8 m/s
so is that the velocity at the instant he throws it?
i seem to be more confused..?:grumpy: :yuck:

Doc Al
Mentor
Would the orginal momentum be zero
so would i use 0 as the momentum and find velocity from there?
Yes, the initial momentum is zero. Before he throws the rock, nothing is moving.
also it says that if oliver throws the rock he would throw it at 19.8 m/s
so is that the velocity at the instant he throws it?
Yes, assume that when he's done throwing the rock, that that is the rock's final speed.

Would the orginal momentum be zero
so would i use 0 as the momentum and find velocity from there?
Yes

also it says that if oliver throws the rock he would throw it at 19.8 m/s
so is that the velocity at the instant he throws it?
Yes. Since no external force is said act in the horizontal direction, it would continue with the same velocity.

Okay, I'm late by only 2 minutes this time. :D

Its seems that since that 19.8 m/s is the velocity oliver will continue with, hence that would be the recoil velocity?   But i tried that its not the answer. Ughhh...:surprised I am gonna think about this question later since i should really study for my bio midterm.:!!)

Doc Al
Mentor
Its seems that since that 19.8 m/s is the velocity oliver will continue with, hence that would be the recoil velocity?
19.8 m/s is the speed of the rock, not Oliver. You have to apply conservation of momentum to solve for Oliver's recoil speed.

I seriously dun get this...

19.8 m/s is the speed of the rock, not Oliver. You have to apply conservation of momentum to solve for Oliver's recoil speed.
Is this wrong?:
(mass of oliver)* (oliver's velocity)= (mass of rock) (velocity of rock)
(59.1)(v)= (0.588) (19.8)
v= 0.588*19.8/59.1
v=0.196?

Is this wrong?:
(mass of oliver)* (oliver's velocity)= (mass of rock) (velocity of rock)
(59.1)(v)= (0.588) (19.8)
v= 0.588*19.8/59.1
v=0.196?
Yes. In the sense that you forgot a minus sign.

From the origianl equation:

0 = msvsf + movof (s refers to stone and o to Oliver, f - the final velocity)

Therfore,
vof = -(msvsf)/mo

The minus sign is important because you are solving for velocity, and not just speed. The sign says that Oliver would move in the opposite direction with a speed of 0.197m/s.

A 60kg teenager on in-line skates intially at rest holds a 0.5kg ball. The teenager throws the ball horizontally at a speed of 12m/s relative to the ground and recoils backwards.

Ignoring friction, what is the recoil speed of the skater?????

Doc Al
Mentor

A 60kg teenager on in-line skates intially at rest holds a 0.5kg ball. The teenager throws the ball horizontally at a speed of 12m/s relative to the ground and recoils backwards.

Ignoring friction, what is the recoil speed of the skater?????
Hint: Momentum is conserved. What's the total momentum of the skater + ball system?

A 60kg teenager on in-line skates intially at rest holds a 0.5kg ball. The teenager throws the ball horizontally at a speed of 12m/s relative to the ground and recoils backwards.

Ignoring friction, what is the recoil speed of the skater?????

Can you please show me in steps???? and what formula do I use?????

Doc Al
Mentor

A 60kg teenager on in-line skates intially at rest holds a 0.5kg ball. The teenager throws the ball horizontally at a speed of 12m/s relative to the ground and recoils backwards.

Ignoring friction, what is the recoil speed of the skater?????

Can you please show me in steps???? and what formula do I use?????
It doesn't work that way. You make an attempt, show what you did and where you are stuck, then we'll jump in and help.

As far as which formula to use: See my hint! And read the thread that you chose to hijack! (It's practically the same problem.)

Use m1*initial velocity1 + m2*initial velocity2= m1*final velocity1+ m2*final velocity2

m1= Oliver
m2= The Rock
Oliver has no initial velocity so initial velocity1 is gone. The rock's finals velocity will be zero, so final velocity2 is gone.
You're left with:

m1+m2*initial velocity2=m1*final velocity1+m2

You should get his recoil velocity from that. If you don't then I'm wrong..and it wouldn't be the first time lol.

(59.1)+(.558)(19.8)=(59.1)*final velocity2+.558